Python：高效计算字典列表中某个键的唯一值数量

Question

在这段Python代码中，我有一个人员列表（人员是字典），我正在尝试查找特定键（在本例中称为Nationality）的唯一值数量（即要查找人员列表中唯一国籍的数量）。希望能有更好的写法：

no_of_nationalities = []
for p in people:
    no_of_nationalities.append(p['Nationality'])
print 'There are', len(set(no_of_nationalities)), 'nationalities in this list.'

非常感谢

- Antony

Answer 1

更好的方法是直接从字典构建set：

print len(set(p['Nationality'] for p in people))

- Sven Marnach

Answer 2

有一个 collections 模块

import collections
....
count = collections.Counter()
for p in people:
    count[p['Nationality']] += 1;
print 'There are', len(count), 'nationalities in this list.'

这样，您也可以统计每个国籍。

print(count.most_common(16))#print 16 most frequent nationalities 

- Nikiton

Answer 3

count = len(set(p['Nationality'] for p in people))
print 'There are' + str(count) + 'nationalities in this list.'

- phihag

---

如果我正确地阅读了原始帖子，people 是一个字典的集合，但不是一个字典本身。 - Sven Marnach

Answer 4

len(set(x['Nationality'] for x in p))

- jonesy